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mart [117]
2 years ago
12

30 POINTS

Mathematics
1 answer:
vivado [14]2 years ago
6 0
Centre= (-1,-2)
Radius=5
All working shown on photo

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P is located at -23 and q is located at 18. what is the best distance between the points
Rina8888 [55]
Distance equals final position minus initial position.

d=18--23

d=18+23

d=41 units
5 0
3 years ago
Read 2 more answers
What is the slope of the line that contains the points 12. 6) and (-1, 6)?
Tema [17]

Answer:

c) 0

Step-by-step explanation:

m= slope

m= (y2 -y1)/(x2-x1)

we have the points:

(12,6) and (-1,6)

x1= 12      y1= 6

x2= -1      y2= 6

so we have:

m= (6-6)/( -1 -12)

m= 0/ -13

m= 0

4 0
3 years ago
A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.
Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

t=\sqrt{\frac{2h}{a}} ,where h=height of release

a=acceleration

a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

Where I=moment of inertia

a for hoop

a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}

a=\frac{g\sin \theta }{2}

a for Uniform solid cylinder

a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}

a=\frac{2g\sin \theta }{3}

a for spherical shell

a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}

a=\frac{3g\sin \theta }{5}

a for Uniform Solid

a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}

a=\frac{5g\sin \theta }{7}

time taken will be inversely proportional to the square root of acceleration

t_1=k\sqrt{2}=1.414k

t_2=k\sqrt{\frac{3}{2}}=1.224k

t_3=k\sqrt{\frac{5}{3}}=1.2909k

t_4=k\sqrt{\frac{7}{5}}=1.183k

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0
3 years ago
The net of a square pyramid is shown:
lys-0071 [83]
1.7625 square inches
3 0
2 years ago
Question 20 of 40
frozen [14]

Answer:

A

Step-by-step explanation:

I guessed also I just took the exam

4 0
3 years ago
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